Question

Determine the enthalpy change of precipitation for the reaction between 50ml of 0.1M sodium sulfate and 50ml of 0.1M barium nitrate, given that the final temperature is 25°C and the initial temperature was 30°C. The specific heat capacity of the solution is 4.18 J/g °C and the density is 1 g/mL. (Assume the reaction goes to completion and all volumes are additive.)

Answer 1

First, let's write the balanced chemical equation for the reaction:Na2SO4  aq  + Ba NO3 2  aq   BaSO4  s  + 2 NaNO3  aq Now, we need to find the limiting reactant. Since both solutions have the same concentration and volume, the limiting reactant will be the one with the smaller stoichiometric coefficient, which is sodium sulfate  Na2SO4  in this case.Next, we'll calculate the moles of the limiting reactant:moles of Na2SO4 =  0.1 mol/L  *  0.050 L  = 0.005 molSince the volumes are additive, the total volume of the solution is:total volume = 50 mL + 50 mL = 100 mLNow, we'll calculate the mass of the solution using the density:mass of solution =  100 mL  *  1 g/mL  = 100 gNow, we can calculate the heat change  q  using the specific heat capacity, mass, and temperature change:q =  mass of solution  *  specific heat capacity  *  T q =  100 g  *  4.18 J/g C  *  25C - 30C q = 100 g * 4.18 J/g C *  -5C q = -2090 JSince the reaction is exothermic, the heat change will be negative. Now, we can calculate the enthalpy change of precipitation  H  by dividing the heat change by the moles of the limiting reactant:H = q / moles of limiting reactantH = -2090 J / 0.005 molH = -418000 J/molThe enthalpy change of precipitation for the reaction between 50 mL of 0.1 M sodium sulfate and 50 mL of 0.1 M barium nitrate is -418 kJ/mol.